Topic 6: Circular motion and gravitation
6.2 – Newton’s law of gravitation
Essential idea: The Newtonian idea of gravitational
force acting between two spherical bodies and the
laws of mechanics create a model that can be used
to calculate the motion of planets.
Nature of science: Laws: Newton’s law of gravitation
and the laws of mechanics are the foundation for
deterministic classical physics. These can be used
to make predictions but do not explain why the
observed phenomena exist.
Topic 6: Circular motion and gravitation
6.2 – Newton’s law of gravitation
Understandings:
• Newton’s law of gravitation
• Gravitational field strength
Applications and skills:
• Describing the relationship between gravitational force
and centripetal force
• Applying Newton’s law of gravitation to the motion of
an object in circular orbit around a point mass
• Solving problems involving gravitational force,
gravitational field strength, orbital speed and orbital
period
• Determining the resultant gravitational field strength
due to two bodies
Topic 6: Circular motion and gravitation
6.2 – Newton’s law of gravitation
Guidance:
• Newton’s law of gravitation should be extended to
spherical masses of uniform density by assuming
that their mass is concentrated at their centre
• Gravitational field strength at a point is the force per
unit mass experienced by a small point mass at that
point
• Calculations of the resultant gravitational field strength
due to two bodies will be restricted to points along
the straight line joining the bodies
Topic 6: Circular motion and gravitation
6.2 – Newton’s law of gravitation
Data booklet reference:
𝑀𝑚
𝐹=𝐺 2
𝑟
𝐹
𝑔=
𝑚
𝑀
𝐹=𝐺 2
𝑟
Theory of knowledge:
• The laws of mechanics along with the law of gravitation create the
deterministic nature of classical physics. Are classical physics
and modern physics compatible? Do other areas of knowledge
also have a similar division between classical and modern in
their historical development?
Topic 6: Circular motion and gravitation
6.2 – Newton’s law of gravitation
Utilization:
• The law of gravitation is essential in describing the
motion of satellites, planets, moons and entire
galaxies
• Comparison to Coulomb’s law (see Physics sub-topic
5.1)
Aims:
• Aim 4: the theory of gravitation when combined and
synthesized with the rest of the laws of mechanics
allows detailed predictions about the future position
and motion of planets
Field (Fundamental) Forces
Gravitational Force
Gravitational force is cumulative and extended to infinity. The gravitational force between two object
is due to the cumulative effect of billions of billions of the atoms made up both bodies. This means
that the larger the body (contain more matter), the stronger the force. But on the scale of individual
particles, the force is extremely small, only in the order of 10-38 times that of the strong force.
Electromagnetic Force
The force is long range, in principle extending over infinite distance. However, the strength can
quickly diminishes due to shielding effect. Many everyday experiences such as friction and air
resistance are due to this force. This is also the resistant force that we feel, for example, when
pressing our palm against a wall. This is originated from the fact that no two atoms can occupy the
same space. However, its strength is about 100 times weaker within the range of 1 fm, where the
strong force dominates. But because there is no shielding within the nucleus, the force can be
cumulative and can compete with the strong force. This competition determines the stability
structure of nuclei.
Weak Nuclear Force
Responsible for nuclear beta decay and other similar decay processes involving
fundamental particles. The range of this force is smaller than 1 fm and is 10-7 weaker than
the strong force
Strong Nuclear Force
This force is responsible for binding of nuclei. It is the dominant one in reactions and decays of most
of the fundamental particles. This force is so strong that it binds and stabilize the protons of similar
charges within a nucleus. However, it is very short range. No such force will be felt beyond the order
of 1 fm (femtometer or 10-15 m).
One of the most significant intellectual achievements in the
history of thought. It is universal – it applies to all objects
regardless of their location anywhere in the Universe. Every
object in the universe attracts every other object.
Gravitational force between two point masses m1 and m2 is
proportional to their product, and inversely proportional to the
square of their separation r.
m1m2
F=G 2
r
r
G = 6.67x10-11 Nm2/ kg2 – “Universal gravitational constant”
the same value anywhere in the universe - very small value
– no significant forces of attraction between ordinary sized objects.
The actual value of G, the universal gravitational constant, was not known
until Henry Cavendish conducted a tricky experiment in 1798 to find it.
Newton’s law of gravitation
The earth, planets,
moons, and even
the sun, have
many
layers – kind of
like an onion:
In other words,
NONE of the
celestial bodies we
observe are point
masses.
Given that the law is called the universal law of
gravitation, how do we use it for planets and such?
Newton’s law of gravitation
Newton spent much time developing integral calculus to prove that
“A spherically symmetric shell of mass M acts as if all of its
mass is located at its center.”
-Newton’s shell theorem.
M
m
r
Thus F = Gm1m2 / r 2 works not only for point masses, which have
no radii, but for any spherically symmetric distribution of mass at
any radius like planets and stars.
 r is the distance between the centers of the masses.
m1
F12
r
F21
m2
FYI The radius of each
mass is immaterial.
EXAMPLE: The earth has a mass of M = 5.981024 kg and the
moon has a mass of m = 7.361022 kg. The mean distance
between the earth and the moon is 3.82108 m. What is the
gravitational force between them?
SOLUTION: F = GMm / r 2.
F = (6.67×10−11)(5.981024 )(7.361022 ) / (3.82108)2
F = 2.011020 N.
What is the speed of the moon in its orbit about earth?
SOLUTION: FC = FG = mv 2 / r .
2.011020 = ( 7.361022 ) v 2 / 3.82108
v = 1.02103 ms-1.
What is the period of the moon (in days) in its orbit about earth?
SOLUTION: v = 2r / T.
T = 2r / v = 2( 3.82108 ) / 1.02103
= (2.35 106 s)(1 h / 3600 s)(1 d / 24 h) = 27.2 d.
Gravitational field strength
Suppose a mass m is located a distance r from a another mass M.
The gravitational field strength g is the force per unit mass acting
on m due to the presence of M.
𝐹
𝑔=
𝑚
The units are newtons per kilogram (N kg -1 = m s -2).
Suppose a mass m is located on the surface of a planet of radius R.
We know that it’s weight is F = mg.
But from the law of universal gravitation, the weight of m is equal to its
attraction to the planet’s mass M and equals F = GMm / R 2.
 mg = GMm / R 2.
𝐺𝑀
𝑔= 2
𝑅
gravitational field strength at surface of
a planet of mass M and radius R
Double the distance from the centre, g is 4 times less,
and so is weight
PRACTICE: A 525-kg satellite is launched from the
earth’s surface to a height of one earth radius above the
surface. What is its weight (a) at the surface, and (b) at
altitude?
SOLUTION:
(a) AT SURFACE: gsurface = 9.8 m s-2.
 F = mg = (525)(9.8) = 5145 N.
W=5100 N
(b) AT ALTITUDE: gsurface+R = 2.45 m s-2.
F = mg = (525)(2.46) = 1286.25 N
W=1300N
Gravitational field strength
Compare the gravitational force formula
F = GMm / r 2 (Force)
with the gravitational field formula
g = GM / r 2
(Field)
Note that the force formula has two masses, and the force is the
result of their interaction at a distance r.
Note that the field formula has just one mass – namely the mass
that “sets up” the local field in the space surrounding it. It “curves”
it.
The field view of the universe (spatial disruption by a
single mass) is currently preferred over the force view
(action at a distance) as the next slides will show.
Consider the force view.
In the force view, the masses know the locations of each
other at all times, and the force is instantaneously felt by both
masses at all times.
This requires the “force signal” to
be transferred between the masses
instantaneously.
 Einstein’s special theory of relativity
states unequivocally that
the fastest any signal can travel
is at the (finite) speed of light c.
SUN
Thus the action at a distance “force signal” will be slightly
mdelayed in telling the orbital mass when to turn.
The end result would have to be an expanding spiral
mmotion, as illustrated in the following animation:
We do not observe planets leaving
mtheir orbits as they travel around the
msun.
Thus force as action at a distance
doesn’t work if we are to believe
special relativity.
 And all current evidence points to
mthe correctness of special relativity.
SUN
So how does the field view take care of this “signal lag”
problem?
Simply put – the gravitational field distorts the space around
the mass that is causing it so that any other mass placed at
any position in the field will “know” how to respond
immediately.
Think of space as a stretched
rubber sheet – like a drum head.
Bigger masses “curve” the rubber
sheet more than smaller masses.
The next slide illustrates this gravitational “curvature” of the
space surrounding, for example, the sun.
 each mass “feels” a different “slope” and must travel
mat a particular speed to stay in orbit.
The field view eliminates the need for long distance
msignaling between two masses. Rather, it distorts the
mspace about one mass.
 In the space surrounding the mass M which sets up
mthe field we can release “test masses” m1 and m2
mas shown to determine the strength of the field.
m1
m2
(a) Because g = GM / r2.
It varies as 1 / r2.
g2
M
g1
(b) Because the
gravitational force
is attractive.
(a) The field arrow is bigger for m2 than m1. Why?
(b) The field arrow always points to M. Why?
By “placing” a series of test masses about a larger
mmass, we can map out its gravitational field:
M
The field arrows of the inner ring are longer than the field arrows of
the outer ring and all field arrows point to the centerline.
Gravitational field strength
 Simplification: Instead of having arrows at every point
we take the convention of drawing “field lines”
as a single arrow.
Density/concentration of the
field lines convey us comparable
strengths.
The closer together the field
lines, the stronger the field.
In the red region the field lines are closer
together than in the green region.
Thus the red field is stronger than the
green field.
SUN
SUN
PRACTICE: Sketch the gravitational field about the earth (a) as
viewed from far away, and (b) as viewed “locally” (at the surface).
SOLUTION:
(a)
(b)
or
 The closer to the surface we are, the more uniform the field
concentration.
EXAMPLE: Find the gravitational field strength at a
point between the earth and the moon that is right
M = 5.981024 kg
between their centers.
m = 7.361022 kg
SOLUTION:
gm
gM
Make a sketch.
d = 3.82108 m
Note that r = d / 2 = 3.82108 / 2 = 1.91 108 m.
gm = Gm / r 2
gm = (6.67×10-11)(7.361022)/(1.91108)2 = 1.3510-4 N.
gM = GM / r 2
gM = (6.67×10-11)(5.981024)/(1.91108)2 = 1.0910-2 N.
Finally, g = gM – gm = 1.0810 -2 N,→.
PRACTICE: Two spheres of equal mass
and different radii are held a
distance d apart. The gravitational
field
strength is measured on the
line joining
the two masses at position x
which varies.
Which graph shows the variation of g with x correctly?
There is a point between M and m where g = 0.
Since g = Gm / R 2 and Rleft < Rright, then gleft > gright at the surfaces
of the masses.
EXAMPLE: Derive Kepler’s law, which states that the period T of an object in a
circular orbit about a body of mass M is given by
2
4𝜋
𝑇2=
𝑟3
𝐺𝑀
SOLUTION:
𝑚𝑣 2
𝑀𝑚
▪ 𝐹𝑐 =
=𝐺 2
𝑟
𝑟
▪
2𝜋𝑟
𝑇
2
𝑀
=𝐺
𝑟
2
4𝜋
▪ 𝑇2 =
𝑟3
𝐺𝑀
𝑀
 𝑣 =𝐺
𝑟
2
EXAMPLE: A satellite in geosynchronous orbit takes 24 hours to
orbit the earth. Thus, it can be above the same point of the earth’s
surface at all times, if desired. Find the necessary orbital radius,
and express it in terms of earth radii. RE = 6.37106 m.
SOLUTION:
𝑇 = (24 ℎ)(3600 𝑠 ℎ−1 ) = 86400 𝑠
𝑚𝑣 2
𝑀𝑚
=𝐺 2
𝑟
𝑟
2𝜋𝑟
𝑇
𝑟3
2

𝑣2
𝑀
=𝐺
𝑟
𝑀
=𝐺
𝑟
𝐺𝑀 2
6.6710−115.981024
3
2
= 2 𝑇 ⇒𝑟 =
(86400)
4𝜋
4𝜋 2
𝑟 = 42250474 𝑚 = 6.63 𝑅𝐸
𝑚𝑣 2
𝑀𝑚
=𝐺 2
𝑟
𝑟
2𝜋𝑟
𝑇
2
=𝐺
𝑀
𝑟
 𝑣2 = 𝐺
⇒ 𝑟3 =
𝑀
𝑟
𝐺𝑀 2
𝑇
2
4𝜋
FYI
Kepler’s third law originally said that the square of the period was
proportional to the cube of the radius – and nothing at all about
what the constant of proportionality was. Newton’s law of
gravitation was needed for that!
𝑚𝑣 2
𝑀𝑚
▪ 𝐹𝑐 =
=𝐺 2
𝑟
𝑟
▪
2𝜋𝑟
𝑇
2
 𝑣2 = 𝐺
𝑀
𝑟
𝑀
=𝐺
𝑟
2
2
4𝜋
4𝜋
▪ 𝑇2 =
𝑟3 ⇒ 𝑇 =
𝐺𝑀
𝐺𝑀
3/2
𝑟 3/2
𝑚𝑣 2
𝑀𝑚
=𝐺 2
𝑟
𝑟
𝑀
 𝑟=𝐺 2
𝑣
𝑀
𝐺 2
𝑟𝑋
𝑣𝑥 𝑣𝑌2
=
=
=
𝑟𝑦 𝐺 𝑀 𝑣𝑥2
𝑣𝑌2
𝑟𝑋
𝑟𝑦
3
2𝜋𝑟𝑌
𝑇𝑌
2𝜋𝑟𝑋
𝑇𝑋
𝑟𝑋
= 64 ⇒
=4
𝑟𝑦
2
𝑇𝑋 𝑟𝑌
=
𝑇𝑌 𝑟𝑋
2
8𝑟𝑌
=
𝑟𝑋
2
Solving problems involving gravitational field
Consider Dobson inside an elevator which is not
moving…
If he drops a ball, it will accelerate downward
at 10 ms-2 as expected.
PRACTICE: If the elevator is accelerating upward at 2 ms-2, what will
Dobson observe the dropped ball’s
acceleration to be?
SOLUTION:
Since the elevator is accelerating upward at 2 ms-2 to meet the ball
which is accelerating downward at 10 ms-2, Dobson would observe an
acceleration of 12 ms-2.
If the elevator were accelerating downward at 2, he would observe an
acceleration of 8 ms-2.
PRACTICE: If the elevator were to accelerate
downward at 10 ms-2, what would Dobson
observe the dropped ball’s acceleration to be?
SOLUTION:
He would observe the acceleration of the ball
to be zero!
He would think that the ball was “weightless!”
FYI
The ball is NOT weightless, obviously. It is
merely accelerating at the same rate as Dobson!
How could you get Dobson to accelerate
downward at 10 ms-2?
Cut the cable!
The “Vomit Comet”
PRACTICE: We have all seen astronauts experiencing
“weightlessness.” Explain why it only appears that they are
weightless.
SOLUTION: The astronaut, the spacecraft, and the tomatoes, are
all accelerating at ac = g.
They all fall together and appear to be weightless.
International Space
Station
PRACTICE: Discuss the concept of weightlessness in deep space.
SOLUTION: Only in deep space – which is defined to be far, far
away from all masses – will a mass be truly weightless.
In deep space, the r in
F = GMm / r 2
is so large for every m
that F, the force of
gravity, is for all
intents and
purposes, zero.
Since the satellite is in circular orbit FC = mv 2/ r.
Since the satellite’s weight is holding it in orbit, FC = mg.
Thus mv 2/ r = mg.
Finally g = v 2/ r.
A satellite of mass m orbits a planet of mass M and radius R as shown.
The radius of the circular orbit of the satellite is x. The planet may be
assumed to behave as a point mass with its mass concentrated at its
centre.
(a) Deduce that the linear speed v of the satellite in its
orbit is given by expression 𝑣 =
𝑣 2 𝐺𝑀
 𝑎𝑐 =
= 2
𝑥
𝑥
𝐺𝑀
𝑔 = 2
𝑥
 𝑣2 =
𝐺𝑀
𝑥
 𝑣=
x
R
𝐺𝑀
𝑥
(ac = g in circular orbits).
𝐺𝑀
𝑥
(b) Derive an expression, in terms of m, G, M and x, for the kinetic energy of the satellite.
 𝐹𝑟𝑜𝑚 𝑎
 𝐸𝐾 =
𝑣2 =
𝐺𝑀
𝑥
1
𝐺𝑀𝑚
𝑚𝑣 2 =
2
2𝑥
This question is about gravitation. A binary star consists of two stars
that each follow circular orbits about a fixed point P as shown. The stars
have the same orbital period T. Each star may be considered to act as a
point mass with its mass concentrated at its centre. The stars, of masses
M1 and M2, orbit at distances R1 and R2 respectively from point P.
(a)
State the name of the force that provides the centripetal force
for the motion of the stars.
 It is the gravitational force.
(b) By considering the force acting on one of the stars, deduce that
the orbital period T is given by the expression
 M1 experiences FC = M1v12/ R1.
 Since v1 = 2R1/ T, then v12 = 42R12/ T 2.
 Thus FC = FG  M1v12/ R1 = GM1M2 / (R1+R2) 2.
M1(42R12/ T 2) / R1 = GM1M2 / (R1+R2) 2
42R1(R1+R2) 2 = GM2T 2

T2
42
=
R (R +R ) 2
GM2 1 1 2
Note that FG = GM1M2 / (R1+R2) 2.
R
M1 1
R2
P
M2
(c) The star of mass M1 is closer to the point P than the star of mass M2.
Using the answer in (b), state and explain which star has the larger mass.
 From (b)
T 2 = (42 / GM2)R1(R1+R2) 2.
 From symmetry
T 2 = (42/ GM1)R2(R1+R2) 2.
(42/ GM2)R1(R1+R2) 2 = (42/ GM1)R2(R1+R2) 2
(1 / M2)R1 = (1 / M1)R2
M1 / M2 = R2 / R1
Since R2 > R1, we see that M1 > M2.
R1
M1
R2
P
M2